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kernel, and we note that (1 - i)(1 + i) = 2. Let m + ni " ker(Æ) = {m + ni |
m + n a" 0 (mod 2)}. Then m and n are either both even or both odd, and
so it follows that m - n is always even. Therefore
m + ni = (m - n) + n + ni = (m - n) + n(1 + i)
m - n
= (1 - i)(1 + i) + n(1 + i)
2
1
= (m - n)(1 - i) + n (1 + i) ,
2
and so m + ni belongs to the principal ideal generated by 1 + i.
(b) For any prime number p, define ¸ : Z[i] ’! Zp[x]/ x2 + 1 by ¸(m + ni) =
[m + nx]. Prove that ¸ is an onto ring homomorphism.
Solution: We have the following computations, which show that ¸ is a ring
homomorphism. We need to use the fact that [x2] = [-1] in Zp[x]/ x2 + 1 .
¸((a + bi) + (c + di)) = ¸((a + c) + (b + d)i) = [(a + c) + (b + d)x]
¸((a + bi)) + ¸((c + di)) = [a + bx] + [c + dx] = [(a + c) + (b + d)x]
¸((a + bi)(c + di)) = ¸((ac - bd) + (ad + bc)i) = [(ac - bd) + (ad + bc)x]
¸((a + bi))Æ((c + di)) = [a + bx][c + dx] = [ac + (ad + bc)x + bdx2] .
Since the elements of Zp[x]/ x2 + 1 all have the form [a + bx], for some
congruence classes a and b in Zp, it is clear the ¸ is an onto function.
16. Let I and J be ideals in the commutative ring R, and define the function
Æ : R ’! R/I •" R/J by Æ(r) = (r + I, r + J), for all r " R.
(a) Show that Æ is a ring homomorphism, with ker(Æ) = I )" J.
CHAPTER 5 SOLUTIONS 99
Solution: The fact that Æ is a ring homomorphism follows immediately from
the definitions of the operations in a direct sum and in a factor ring. Since
the zero element of R/I •" R/J is (0 + I, 0 + J), we have r " ker(Æ) if and only
if r " I and r " J, so ker(Æ) = I )" J.
(b) Show that if I + J = R, then Æ is onto, and thus R/(I )" J) R/I •" R/J.
=
Solution: If I +J = R, then we can write 1 = x+y, for some x " I and y " J.
Given any element (a + I, b + J) " R/I •" R/J, consider r = bx + ay, noting
that a = ax + ay and b = bx + by. We have a - r = a - bx - ay = ax - bx " I,
and b-r = b-bx-ay = by -ay " J. Thus Æ(r) = (a+I, b+J), and Æ is onto.
The isomorphism follows from the fundamental homomorphism theorem.
17. Considering Z[x] to be a subring of Q[x], show that these two integral domains
have the same quotient field.
f(x)
Solution: An element of the quotient field of Q[x] has the form , for
g(x)
polynomials f(x) and g(x) with rational coefficients. If m is the lcm of the
denominators of the coefficients of f(x) and n is the lcm of the denominators
f(x) h(x)
n
of the coefficients of g(x), then we have = for h(x), k(x) " Z[x],
g(x) m k(x)
f(x)
and this shows that belongs to the quotient field of Z[x].
g(x)
18. Let p be an odd prime number that is not congruent to 1 modulo 4. Prove
that the ring Zp[x]/ x2 + 1 is a field.
Hint: Show that a root of x2 = -1 leads to an element of order 4 in the
multiplicative group Z×.
p
Solution: We must show that x2 + 1 is irreducible over Zp, or, equivalently,
that x2 + 1 has no root in Zp.
Suppose that a is a root of x2 + 1 in Zp. Then a2 a" -1 (mod p), and so
a4 a" 1 (mod p). The element a cannot be a root of x2 - 1, so it does not have
order 2, and thus it must have order 4. By Lagrange s theorem, this means
that 4 is a divisor of the order of Z×, which is p - 1. Therefore p = 4q + 1 for
p
some q " Z, contradicting the assumption.
100 CHAPTER 5 SOLUTIONS
Chapter 6
Fields
SOLUTIONS TO THE REVIEW PROBLEMS
1. Let u be a root of the polynomial x3 +3x+3. In Q(u), express (7-2u+u2)-1
in the form a + bu + cu2.
Solution: Dividing x3 + 3x + 3 by x2 - 2x + 7 gives the quotient x + 2
and remainder -11. Thus u3 + 3u + 3 = (u + 2)(u2 - 2u + 7) - 11, and so
(7 - 2u + u2)-1 = (2 + u)/11 = (2/11) + (1/11)u.
" "
2. (a) Show that Q( 2 + i) = Q( 2, i).
" " "
Solution: Let u = 2 + i. Since ( 2 + i)( 2 - i) = 2 - i2 = 3, we have
" " "
2 - i = 3( 2" i)-1 " Q(u), and it follows easily that 2 " Q(u) and
+
i " Q(u), so Q( 2, i) †" Q(u). The reverse inclusion is obvious.
"
(b) Find the minimal polynomial of 2 + i over Q.
" " " "
Solution: We have Q †" Q( 2) †" Q( 2, i). Thus [Q( 2) : Q] = 2 since 2
"
is a of a polynomial of degree 2 but is not in Q. We have [Q( 2, i) :
"root "
Q( 2)] =" since i is a root of a polynomial of degree 2 over Q( 2) but is
2
"
not in Q( 2). Thus [Q( 2 + i) : Q] = 4, and so the minimal polynomial for
"
2 + i must have degree 4.
" "
Since u = 2 + i, we have u - i = 2, u2 - 2iu + i2 = 2, and u2 - 3 = 2iu.
Squaring again and combining terms gives u4 -2u2 +9 = 0. Thus the minimal
"
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